Factor the following expression: $x^2 + 6x - 16$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {6}x& & {-16} \end{eqnarray} $ The coefficient on the $x$ term is $6$ and the constant term is $-16$ , so to reverse the steps above, we need to find two numbers that add up to $6$ and multiply to $-16$ You can try out different factors of $-16$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {6}$ $ {a} \times {b} = {-16}$ The two numbers $8$ and $-2$ satisfy both conditions: $ {8} + {-2} = {6} $ $ {8} \times {-2} = {-16} $ So we can factor the expression as: $(x + {8})(x {-2})$